[kimyoung] WEEK 02 Solutions

counting-bits/kimyoung.js

@@ -0,0 +1,24 @@
+var countBits = function (n) {
+  let result = [];
+  for (let i = 0; i <= n; i++) {
+    result.push(countOneBits(i));  // count one bits for each i until n
+  }
+
+  function countOneBits(num) { // method to count one bits
+    let oneBitCount = 0;
+    while (num) {
+      oneBitCount += num % 2;
+      num = num >> 1;
+    }
+    return oneBitCount;
+  }
+
+  return result;
+};
+
+// test cases
+console.log(countBits(2)); // [0,1,1]
+console.log(countBits(5)); // [0,1,1,2,1,2]
+
+// time - O(nlogn) 
+// space - O(n)

encode-and-decode-strings/kimyoung.js

@@ -0,0 +1,47 @@
+/**
+ * Encodes a list of strings to a single string.
+ *
+ * @param {string[]} strs
+ * @return {string}
+ */
+
+const DELIMITER = "#";
+
+var encode = function (strs) {
+  let temp = "";
+  for (const str of strs) { // encode using the delimiter by attaching the length of the array as well
+    temp += `${str.length}${DELIMITER}` + str; 
+  }
+  return temp;
+};
+
+/**
+ * Decodes a single string to a list of strings.
+ *
+ * @param {string} s
+ * @return {string[]}
+ */
+var decode = function (s) { // decode using the length of array that is attached with the delimiter
+  let result = [];
+  let i = 0;
+  while (i < s.length) {
+    let j = i;
+    while (s[j] !== DELIMITER) j++;
+    const len = Number(s.slice(i, j));
+    const word = s.slice(j + 1, j + len + 1);
+    result.push(word);
+    i = j + len + 1;
+  }
+  return result;
+};
+
+/**
+ * Your functions will be called as such:
+ * decode(encode(strs));
+ */
+
+// test cases
+console.log(["Hello", "World"]);
+
+// time - O(n) - iterate through the list of strings once
+// space - O(n) - stores the strings

valid-anagram/kimyoung.js

@@ -0,0 +1,40 @@
+// First Approach - hash map
+var isAnagram = function (s, t) {
+  let map = {};
+  for (const char of s) {
+    // count character occurence of s
+    map[char] ? map[char]++ : (map[char] = 1);
+  }
+  for (const char of t) {
+    // compare character occurence of t to the map object
+    if (map[char]) {
+      map[char]--; // decrement each time
+    } else {
+      return false; // if there's a new character, return false
+    }
+  }
+  for (const el of Object.values(map)) {
+    // if all the values of the map object is 0, return true
+    if (el !== 0) return false; // otherwise return false;
+  }
+  return true;
+};
+
+// test cases
+console.log(isAnagram("anagram", "nagarma"));
+console.log(isAnagram("rat", "car"));
+
+// time - O(s + t) - iterate through both input strings
+// space - O(n) - map obj
+
+//Second Approach - sorted strings
+var isAnagram = function (s, t) {
+  return s.split("").sort().join("") === t.split("").sort().join("");
+};
+
+// test cases
+console.log(isAnagram("anagram", "nagarma"));
+console.log(isAnagram("rat", "car"));
+
+// time - O(nlogn) - using sort method
+// space - O(1) - no extra space memory